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Question Bank (3 marks) · Q18

Q.Show how an operational amplifier can be used as a perfect adder.

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[!TLDR]

A summing amplifier with all resistors equal is a perfect adder: Vo=−(V1+V2+V3)V_o = -(V_1+V_2+V_3).

Circuit: inputs V1,V2,V3V_1, V_2, V_3 are applied through equal resistors R1=R2=R3=RR_1 = R_2 = R_3 = R to the inverting (−) summing node; an equal feedback resistor Rf=RR_f = R connects output to that node; the (+) input is grounded.

Derivation: The summing node is a virtual ground, so the input currents V1/RV_1/R, V2/RV_2/R, V3/RV_3/R add and flow through RfR_f:

Vo=−Rf(V1R+V2R+V3R)V_o = -R_f\left(\frac{V_1}{R}+\frac{V_2}{R}+\frac{V_3}{R}\right)

With Rf=RR_f = R all the weighting coefficients become unity, so

Vo=−(V1+V2+V3)V_o = -(V_1 + V_2 + V_3) …

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